# Binomial and normal distros

1. Sep 24, 2011

### RedX

I want to show that the binomial distribution:

$$P(m)=\frac{n!}{(n-m)!m!}p^m(1-p)^{n-m}$$

using Stirling's formula:

$$n!=n^n e^{-n} \sqrt{2\pi n}$$

reduces to the normal distribution:

$$P(m)=\frac{1}{\sqrt{2 \pi n}} \frac{1}{\sqrt{p(1-p)}} exp[-\frac{1}{2}\frac{(m-np)^2}{np(1-p)}]$$

Unfortunately, I keep on getting an extra term linear in m-np:

$$exp[\frac{m-np}{2n}\frac{2p-1}{(1-p)p}]$$

This term is zero if p=1/2, but I want to show that the binomial distribution reduces to the normal distribution for any probablity p.

Has anyone else had this problem, as I'm sure this derivation is fairly common?

2. Sep 24, 2011

### mathman

It would help to show your derivation in detail. In any case the central limit theorem could be used.

3. Sep 24, 2011

### RedX

Sure. It's a little bit lengthy though, so it might take some work to read it:

$$P(m)=\frac{\sqrt{2 \pi n}n^ne^{-n}}{\sqrt{2 \pi m}m^me^{-m}*\sqrt{2 \pi (n-m)}(n-m)^{(n-m)}e^{-(n-m)}}p^m(1-p)^{n-m}= \frac{n^{n+1}}{\sqrt{2 \pi n}*m^{m+\frac{1}{2}}*(n-m)^{(n-m)+\frac{1}{2}}}p^m(1-p)^{n-m}$$

$$\frac{n^{n+1}}{\sqrt{2 \pi n}*m^{m+\frac{1}{2}}*(n-m)^{(n-m)+\frac{1}{2}}}p^m(1-p)^{n-m}= \frac{1}{\sqrt{2\pi n}} \exp[\log[\frac{n^{n+1}}{m^{m+\frac{1}{2}}*(n-m)^{(n-m)+\frac{1}{2}}}p^m(1-p)^{n-m}]]$$

$$\frac{1}{\sqrt{2\pi n}} \exp(\log[\frac{n^{n+1}}{m^{m+\frac{1}{2}}*(n-m)^{(n-m)+\frac{1}{2}}}p^m(1-p)^{n-m}]]= \frac{1}{\sqrt{2\pi n}}\exp[(n+1) \log n -(m+^1/_2) \log m$$

$$-(n-m+^1/_2) \log (n-m)+m\log p + (n-m) \log[1-p] ]$$

Now I set m=pn+k in the expression, where pn is the average number of successes, and k is the difference from the average, and use that k<<<pn to expand the logarithms in power series up to second order (i.e., ln(1+x)=x-.5x^2).

So for example:

$$(m+^1/_2) \log m=(pn+k+^1/_2)\log (pn+k)= (pn+k+^1/_2)[\log (pn)+\log (1+k/(pn))]= (pn+k+^1/_2)[\log (pn)+k/(pn)-.5 k^2/(pn)^2]$$

After keeping all terms up to second order, I get:

$$P(m)=\frac{1}{\sqrt{2 \pi n}}\frac{1}{\sqrt{p(1-p)}}\exp[-\frac{1}{2}\frac{k^2}{np(1-p)}] \exp[\frac{k}{2n} \frac{2p-1}{(1-p)p}]$$

and that last factor doesn't make sense, but I've checked my results three times and it keeps coming out. But if p=1/2, then it becomes 1, and I do get the normal distribution.

4. Sep 24, 2011

### Stephen Tashi

Just a thought: Stirlings approximation for n! doesn't converge to n! as n approaches infinity. However the ratio of the approximation to n! converges to 1. Does any one know the situation whe we use Stirings approximations for $C_r^n$ ?

5. Sep 26, 2011